Lecture Note on Linear Algebra 15. Dimension and Rank

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1 Lecture Note on Linear Algebra 15. Dimension and Rank Wei-Shi Zheng, 211 November 1, What Do You Learn from This Note We still observe the unit vectors we have introduced in Chapter 1: 1 e 1 =, e 2 = 1, e 3 =. (1) 1 We know the above are the basis (specially the standard basis) of R 3. However, we still have to answer the following question: Question: Why there are three basis in R 3? Basic Concept:dimension( 维 数 ), rank( 秩 ) 1

2 2 Dimension Theorem 1. Let V be a vector space with basis B = { b 1,..., b n }. Then any subset of V containing more than n vectors is linearly dependent. Proof. Let p N and p > n. Assume that { u 1,..., u p } V. As B is the basis, so we have for each u i there exists a coefficient vector a i R n such that u i = B a i, B = [ b 1,..., b n ]. Let A = [ a 1,..., a p ], U = [ u 1,..., u p ]. Then U = BA. The if there is a series of weight c 1,, c p such that c 1 u c p u p =, that is U c =, then we have BA c =. As the number of rows is smaller than the number of columns of BA, so that is there are non-pivot columns in BA. This leads to the matrix equation BA c has non-trivial solution. So there exists non-zeros c such that U c =. That is u 1,..., u p are linearly dependent. Theorem 2. Let V be a vector space with bases B, C of sizes m, n N respectively. Then m = n. That is every basis of V has the same size. Proof. Since C is linearly independent, by Theorem 1, we must have m n. Similarly, n m. So m = n. Definition 3 (dimension). Let V be a vector space with a finite subset as its basis. Then the size of its basis is called the dimension of V and is denoted by dim V, and we say V is finite-dimensional. Otherwise, if V can not be spanned by a finite set then V is infinite-dimensional. Remark: (1) dim R n = n. (2) The space with all polynomials ( 多 项 式 空 间 ) is infinite-dimensional. 2

3 3 Subspace & Dimension Theorem 4. Let V be a vector space of dimension n N. Then any linearly independent subset { v 1,..., v m } of V can be extended to a basis of V. Proof. We prove this theorem by the following steps: 1. When m = n, because if Span{ v 1,..., v m } = V then { v 1,..., v m } is a basis of V. 2. When m < n Span{ v 1,..., v m } V and we can pick v m+1 V Span{ v 1,..., v m }. We claim that { v 1,..., v m, v m+1 } is linearly independent. Assume this is not the case, then there exist c 1,..., c m, c m+1 which are not all zero such that c 1 v c m v m + c m+1 v m+1 =. If c m+1 = then { v 1,..., v m } is linearly dependent, otherwise v m+1 Span{ v 1,..., v m }. Both cases contradict the assumptions we made. So { v 1,..., v m, v m+1 } is linearly independent. 3. If { v 1,..., v m, v m+1 } spans V then we obtain a basis, otherwise repeat this process again and finally, we must obtain a linearly independent set containing { v 1,..., v m } spans V, that is a basis. Theorem 5. Let H be a subspace of a finite-dimension vector space V. Any linearly independent set in H can be expanded, if necessary, to a basis for H. Also, H is finite-dimensional and dimh dimv Theorem 6. Let V be a vector space of dimension p( 1). Any linearly independent set of exactly p elements in V is automatically a basis for V. Any set of exactly p elements that spans V is automatically a basis for V. Proof. By theorem 4, any linearly independent set S of size q( p) can be expanded to the basis of V. This implies, if q = p, then S must be the basis of V. Theorem 7. The dimension of NulA is the number of free variables in the equation A x =, and the dimension of ColA is the number of pivot columns in A. 3

4 4 Rank of a Matrix A 4.1 Row Space Definition 8 (Row Space). If A is an m n matrix. The set of all linear combinations of the row vectors is called the row space of A, denoted by RowA. From another point of view, the rows of A are identical to the columns of A T, so the row space of A can also be written as ColA T. Theorem 9. If two m n matrices A and B are row equivalent, their row spaces are the same. Proof. Since A and B are row equivalent, there exists a invertible matrix G such that B = GA. That is B T = A T G T = A T F, F = G T. By using the matrix partition theory, let A T = [Col 1 A T,, Col m A T ], F = [Col 1 F,, Col m F ], then That is [Col 1 B T,, Col i B T,, Col m A T ] =B T =A T F =[A T Col 1 F, A T Col i F,, A T Col m F ]. Col i B T =A T Col i F =[Col 1 A T,, Col m A T ]([F ] 1i,, [F ] mi ) T =[F ] 1i Col 1 A T + + [F ] mi Col m A T. Since (Row i A) T = Col i A T and (Row i B) T = Col i B T, so Row i B = [F ] 1i Row 1 A + + [F ] mi Row m A. That is RowB RowA. Conversely, by changing the roles of A and B in the above, we can also have RowB RowA. So, RowB = RowA. Theorem 1. If two m n matrices A and B are row equivalent, and B is in echelon form, then the nonzero rows of B form a basis for the row space of A as well as for that of B (2) (3) 4

5 4.2 Definition and Properties of Rank Definition 11 (column & row rank). Let A be a m n matrix. The column rank of A is defined to be dim Col(A) and the row rank of A to be dim Row(A). Remark: The dimension of the null space is sometimes called the nullity of A. Theorem 12 (Rank Theorem). The dimensions of the column space and the row space of an m n matrix A are equal. Also, it holds that: ranka + dim NulA = n. (4) Proof. We know (1) ranka is the number of pivot columns in A. That is if B is an echelon form of A. i.e. ranka is the number of pivot positions in B. (2) Each pivot position corresponds to a nonzero row, and these rows form a basis for the row space of A, so the rank of A is also the dimension of the row space. (3) The dimension of NulA equals the number of free variables in the equation A x =. That is the dimension of NulA is the number of columns of A that are not pivot columns. (4) Number of pivot columns + number of non-pivot columns = number of columns, i.e. ranka + dim NulA = n. 5

6 5 Computation of Null Space and Rang S- pace We are now discussing how to compute bases of Col(A) and Nul(A) given A R m n. Computation of Column Space: Let A = ( a 1 a n ) R m n. Then there exists an invertible matrix B R m m such that BA is in REF. Let B a i1,..., B a ir be all the pivot columns of BA. Then it is obvious that {B a i1,..., B a ir } is a basis of Col(BA). Since B is invertible and by Theorem 15.4, { a i1,..., a ir } forms a basis of Col(A), that is the set of pivot columns of A is a basis of Col(A). Computation of Null Space: We also know that Nul(A) is exactly the solution set of the equation A x =. So by solving Ax = using Row Reduction Algorithm, we obtain v 1,..., v n r Nul(A), such that Nul(A) = {x j1 v x jn r v n r x j1,..., x jn r R} = Span{ v 1,..., v n r }, where x j1,..., x jn r correspond to the free variables of Ax =. But by the Rank Theorem, dim Nul(A) = n dim Col(A) = n r. Thus { v 1,..., v n r } is a basis of Nul(A) by Theorem 5. Examples: Textbook P.24, P.241, P

7 6 Rank & Matrix Inverse Theorem 13 (The Invertible Matrix Theorem). Let A be an n n matrix. The following statements are each equivalent to the statement that A is an invertible matrix: (1) The columns of A form a basis of R n ; (2) ColA = R n ; (3) dim ColA = R n ; (4) ranka = n; (5) NulA = { }; (6) dim NulA = ; Flora, by Titian 7

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